Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is the algebraic intuition behind Vieta jumping in IMO1988 Problem 6? Problem 6 of the 1988 International Mathematical Olympiad notoriously asked:
Let $a$ and $b$ be positive integers and $k=\frac{a^2+b^2}{1+ab}$. Show that if $k$ is an integer then $k$ is a perfect square.
The usual way to show this involves a ... | I don't have a strong math background, I think my solution is more suitable for the Math olympiad contest as a young student.
Problem Definition:
Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$. Show that
$\frac{a^2+b^2}{ab+1}$
is the square of an integer.
Proof:
Let $ t = \frac{a^2+b^2}{ab+1}$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "104",
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Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it ... | Another way to split the integrals is to replace the square root by a power $p$:
$$\sqrt{x^4 + a^4}\longrightarrow \left(x^4 + a^4\right)^{p}$$
The integral of the separate terms will then converge for $p<-\frac{1}{4}$ and can be expressed in terms of the beta-function. You can substitute $p = \frac{1}{2}$ in the final... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the firs... | To test your hypothesis you could work out the form of the differences from the first few cases.
\begin{align*}
1^{3}-0^{3}&=1\\
2^{3}-1^{3}&=7\\
3^{3}-2^{3}&=19\\
4^{3}-3^{3}&=37
\end{align*}
For example rewrite out $(37-19)-(19-7)=18-6=6$ as:
\begin{align*}
\{(4^{3}-3^{3})-(3^{3}-2^{3})\}&-\{(3^{3}-2^{3})-(2^{3}-1^{3... | {
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BMO2 1995 Question 1 - Solve the equation for positive integers (a,b,c) Find all triples of positive integers (a, b, c) such that
$$ \left( 1 + \frac{1}{a} \right)\left( 1 + \frac{1}{b} \right)\left( 1 + \frac{1}{c} \right) = 2 $$
I tried to expand, yielding $ a+b+c+ab+bc+ca+1 = abc $, but I couldn't go anywhere from h... | Considering cases works pretty well. If the product is to be $2$, then at least one of the numbers must be less than $4$, and all numbers are clearly $>1$.
Order the numbers so that $a\le b\le c$. Then $a = 2$ or $a=3$. If $a=2$, then $4\le b\le 6$ (since $b=2$ or $b=3$ produces a product too large, and if $b=7$ then $... | {
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Integrals $\int\limits _{0}^{\pi/4}\frac{dx}{1+\cos^2x},~~\int\limits_{0}^{\pi/4}\frac{1-\cos^2x}{1+\cos^2x}dx$
Evaluate the definite integrals:
$$I_1=\int\limits _{0}^{\pi/4}\frac{dx}{1+\cos^2x},~~I_2=\int\limits_{0}^{\pi/4}\frac{1-\cos^2x}{1+\cos^2x}dx.$$
Thanks for the help.
Attempt. The classic substitution $\tan... | HINT:
*
*Substitute $u=\tan(x)$ and $\text{d}u=\sec^2(x)\space\text{d}x$:
$$\int\frac{1}{1+\cos^2(x)}\space\text{d}x=\int\frac{\sec^2(x)}{1+\sec^2(x)}\space\text{d}x=\int\frac{\sec^2(x)}{2+\tan^2(x)}\space\text{d}x=\int\frac{1}{2+u^2}\space\text{d}u$$
*Substitute $s=\cot(x)$ and $\text{d}s=-\csc^2(x)\space\text{d}x... | {
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Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| By Vieta's formulas $x$ and $y$ are roots of the polynomial
$$ p(z)=z^2-3z+3 $$
and if $z\in\{x,y\}$ we have $z^2=3z-3$, from which $z^4=9z^2-18z+9=9z-18$ and
$$ x^4+y^4 = 9(x+y)-36 = \color{red}{-9}.$$
In general, the characteristic polynomial of the sequence $\{s_n=x^n+y^n\}_{n\geq 0}$ is exactly $p(z)$,
so $s_{n+2}... | {
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Proof Verification : Prove -(-a)=a using only ordered field axioms I need to prove for all real numbers $a$, $-(-a) = a$ using only the following axioms:
Thanks to many members of the Mathematics Stackexchange Community, I have the following proof worked out:
Theorem: The Additive Inverse Identity is Unique
$
( \fora... | An extended comment:
*
*If you want a proof verification it make sense that you number your equations so that they are easy to reference. You can use \$\tag{1}\$ in the equation code and reference it as \$(1)\$.
*Start end end your LaTeX blocks wiht \$\$ and not with \$.
Lets make
$$
\begin{align}
a \cdot 0 + a ... | {
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Proof related to Fibonacci sequence Could anyone help me with this problem?
$$
\sum_{j=0}^{n}\binom{n}{j}F_{n+1-j}= F_{2n+1}
$$
I used induction and was able to get to this:
$$
2\sum_{j=0}^{n}\binom{n}{j}F_{n-j}= F_{2n}
$$
However I still do not know how to prove the second equality. Really appreciate any help
| (For future reference as an exercise in integration.) Suppose we seek
to verify that
$$F_{2n} = \sum_{j=0}^n {n\choose j} F_{n-j}$$
with $F_q$ being a Fibonacci number. These have generating function
$$f(z) = \frac{z}{1-z-z^2}$$
and hence
$$F_{n-j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-j+1}} \frac{z}{1-... | {
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Maximum value of trigonometric equation $\cos^2(\cos\theta) + \sin^2(\sin\theta)$ For any real $\theta$ the maximum value of $$\cos^2(\cos\theta) + \sin^2(\sin\theta)$$
A. $1$
B. $1 + \sin^21$
C. $1 + \cos^21$
D. does not exist
I tried it by converting the whole expression into $\sin$ but getting nowhere with that.
$$... | $$\max_{\theta\in\mathbb{R}}\{\cos^2(\cos\theta) + \sin^2(\sin\theta)\}\le\max_{-1\le x \le 1}\cos^2 x+\max_{-1\le y \le 1}\sin^2 y=1+\sin^21$$
$$\cos^2 \left( \cos \frac{\pi}{2} \right) + \sin^2 \left( \sin \frac{\pi}{2} \right) = 1 + \sin^2 1$$
So, the maximum is at most $1+\sin^21$, and this value is achieved. Hence... | {
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How to calculate this imaginary part of a complex square root? The calculation of the square root of a complex number $a + ib$ involves solving the equation
$$ (x + iy)^2 = a + ib$$
So far so good. One obtains the equations
$$ 4x^4 -4ax^2 - b^2 = 0, y = b/2x$$
and using the quadratic formula for $x^2$ one gets
$$ x = ... | *
*When $\text{a}\space\wedge\space\text{b}\in\mathbb{C}$:
$$\text{a}=\text{b}^2=\left(\Re[\text{b}]+\Im[\text{b}]i\right)^2=\Re^2[\text{b}]-\Im^2[\text{b}]+2\Re[\text{b}]\Im[\text{b}i]$$
*Solving $x$:
$$y=\frac{\text{b}}{2x}\Longleftrightarrow\text{b}=2xy\Longleftrightarrow x=\frac{\text{b}}{2y}$$
*Solving $x$ by s... | {
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Calculus 2: Partial fractions problem. Finding the value of a constant I encountered the following problem.
Let $f(x)$ be a quadratic function such that $f(0) = -6$ and
$$\int \frac{f(x)}{x^2(x-3)^8} dx $$
is a rational function.
Determine the value of $f'(0)$
Here's what I tried. I decomposed the fraction integrand b... | Note that since $f$ is quadratic with $f(0)=0$, we can write $f(x)$ as $$f(x)=-6+f'(0)x+\frac12 f''(0)x^2$$
Then, the integrand becomes
$$\frac{f(x)}{x^2(x-3)^8}=\frac{-6}{x^2(x-3)^8}+\frac{f'(0)}{x(x-3)^8}+\frac{\frac12 f''(0)}{(x-3)^8} \tag 1$$
The last term on the right-hand side of $(1)$ integrates to the rationa... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $ My answer: $2^\frac{-1}{\sqrt{2}} < x < 1$
Textbook answer: $2^\frac{-12}{17} < x < 1$
The only difference between my resolution and the Textbook one is that I solved by saying that
$$\text{(I) }\sqrt{1 - 8\cdot (\log_{1/4}{x})^2} > 1 -3\cdot \log_{... | at first we have $$x>0$$ and $$\frac{1}{8}\geq \frac{(\ln(x))^2}{(2\ln(2))^2}$$
this condition gives us
$$e^{\ln(2)/\sqrt{2}}\geq x\geq1$$
or
$$e^{-\ln(2)/\sqrt{2}}\le x<1$$
now we can solve the inequality
$$1+\frac{3\ln(x)}{2\ln(2)}<\sqrt{\left(1-\frac{2\ln(x)^2}{\ln(2)^2}\right)}$$
and we do case work:
I)if $$x\le e... | {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim
Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$
For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt ... | Since you already have other answers, I will instead offer a small generalization.
Suppose $p=\frac{k^n +1}{k+1}$ is a prime number. Then, $p$ is the largest prime factor of $\binom{k^n}{2}-1$ if $k=2j^2$ for some integer $j$.
Proof:
We have:
$$\binom{k^n}{2}-1 = \frac{1}{2}k^n(k^n-1)-1 = \frac{1}{2}(k^{2n}-1)-\frac{1}... | {
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Finding the value of $x+y+z$ given sums of two and square root of the other Given: $$x+y+\sqrt{z}=148$$
$$x+\sqrt{y}+z=82$$
$$\sqrt{x}+y+z=98$$
Find the value of $x+y+z$ such that x, y, and z are positive integers.
The only significant step I did was to let $a^2=x, b^2=y, c^2=z.$ After that, I just played with the equa... | Following on from your step, we subtract the second equation from the third. We then have
$$a-a^2 +b^2-b =16$$
or
$$(b-a)(b+a-1)=16$$
No the possible combinations are: $\{1,16\}$, $\{2,8\}$, or $\{4,4\}$. A quick check tells you only the first will work. Thus $b-a=1$ and $b+a-1= 16$. From this we get $a=8$ and $b=9$. ... | {
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$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| We have
$$a^4\le a^4+b^4+c^4+d^4=16\implies a\le 2$$
So, we can have $a^4(a-2)\le 0$, i.e.
$$a^5\le 2a^4$$
Similarly,
$$b^5\le 2b^4,\quad c^5\le 2c^4,\quad d^5\le 2d^4$$
giving
$$a^5+b^5+c^5+d^5\le 2(a^4+b^4+c^4+d^4)=32$$
Note here that
$$a^5+b^5+c^5+d^5=2(a^4+b^4+c^4+d^4)$$
holds when
$$a^4(a-2)=b^4(b-2)=c^4(c-2)=d^... | {
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Intermediate Modular Arithmetic Strategy And Questions I have these problems and I know how to solve ones with two equations but not more. Can anyone provide solutions? Thanks in advance!
Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 1 \pmod{7}, \\
N &\equiv 7 \... | For problem 1:
$$N=7p+1$$
$$7p+1 \equiv 7 \pmod{13}$$
$$7p \equiv 6 \pmod{13}$$
$$p \equiv 12 \pmod{13}$$
(This is calculated by finding the inverse of 7 mod 13).
$$N \equiv 7*12+1 \pmod{13*7}$$
$$N \equiv 85 \pmod{91}$$
$$N = 91q + 85$$
$$N \equiv 13 \pmod{20} $$
$$91q + 85 \equiv 13 \pmod{20} $$
$$91q \equiv 8 \pmod{... | {
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Evaluation of Trigonometric Integral
Evaluation of $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx$
$\bf{My\; Try::} $ Let $$I= \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{(\sin x+\cos x)(2-\... | Hint
From where you left off:
The first integrand is the same as $$-\csc t-4\sin t+4\sin^3t$$
The second integral will be evaluated using the substitution $u=\cos t$ followed by a dose of partial fractions.
| {
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Find the probability of the event that $c=a+b$
In a set $\{0,1,2,3,\dots,n\}$, $3$ numbers $(a,b,c)$ are randomly chosen.
What is the probability of the event that $c=a+b$?
I thought, I should start with $c=0$ for which there is only $1$ way $c=0$, if both $a$ and $b$ are $0$.
For $c=1, 2$ ways: $a=1$ and $b=0$ or $a... | You pointed out that given $c$, the number of ways to have $a+b=c$ is $c+1$. We now must figure out how many total ways there are to have $a+b=c$. This is a simple sum over the range of $c$:
$$\sum_{c=0}^n c+1$$
$$n+1 +\sum_{c=0}^n c$$
$$n+1 +\frac{n(n+1)}{2}$$
$$\frac{(n+2)(n+1)}{2}$$
To find the probability of this e... | {
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What is the sum of all positive even divisors of 1000? I know similar questions and answers have been posted here, but I don't understand the answers. Can anyone show me how to solve this problem in a simple way? This is a math problem for 8th grade students.Thank you very much!
What is the sum of all positive even di... | $n$ is a positive even divisor of $1000$ if and only if $n = 2m$ where $m$ is a divisor of $500$. Since $500 = 2^2 \times 5^3$, there are $(2+1)(3+1) = 12$ divisors of $500$. Those divisors are
\begin{array}{rr}
1, & 500, \\
2, & 250, \\
4, & 125, \\
5, & 100, \\
10, & 50, \\
20, & 25 \\
\end{ar... | {
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Calculus - Finding limit (NOT L'Hopital's Rule): $\lim_{x \to 1^-}\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}$ How do I find this limit?
$$\displaystyle{\lim_{x \to 1^-}}\frac{x^2+x+\sin({\pi \over 2}x)-3}{x-1}$$
I am unable to factor the numerator to get rid of the denominator. Can someone please help? Thank you!
Is there... | Another possible way to do it.
Start changing variable $x=y+1$; this gives
$$\frac{x^2+x+\sin \left(\frac{\pi x}{2}\right)-3}{x-1}=\frac{y^2+3 y+\cos \left(\frac{\pi y}{2}\right)-1}{y}$$ Now, use Taylor expansion $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ which gives $$\cos \left(\frac{\pi y}{2}\right)=1-\frac{\p... | {
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When does this "primes" matrix have full row and column rank? When does the following type of matrix have full row and column rank?
For each entry, we take the value $v$=row*column.
Then for the $b$th bit of $v$, $v_b$ we take the $b$th prime to the $v_b$ power.
EXAMPLES
Say row = 5 and column = 7.
This is 5*7=35, whic... | Easy counterexample, take $N = 3$ and you will see that it is not full ranked:
$\left( \begin{array}{ccc}
2^03^05^0=1 & 2^03^05^0=1 & 2^03^05^0=1 \\
2^03^05^0=1 & 2^13^05^0=2 & 2^03^15^0=3 \\
2^03^05^0=1 & 2^03^15^0=3 & 2^03^05^1=5 \end{array} \right)$, this matrix is not full rank, so your question's answer would be ... | {
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Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - ... | For the first part-
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} $$
$$=\frac{(\operatorname{cosec}B-\cot B)^2+(\operatorname{cosec}B-\cot B)^2}{1}$$
$$=2(\operatorname{cosec^2}B+\cot^2B)$$
$$=2(2\operatorname{... | {
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Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality:
$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$
I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\le... | Let $y=2\sin2x+3\cos^2x=\cos^2x(4\tan x+3)$
$$\iff y\tan^2 x-4\tan x+y-3=0$$ which is a Quadratic Equation in $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
i.e., $$4^2-4y(y-3)=-4(y+1)(y-4)\ge0\iff(y+1)(y-4)\le0$$
$$\iff-1\le y\le4$$
| {
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"answer_id": 3
} |
Finding $x$ such that $2x^2-2x+1=(2m+1)^2$ We need to find the smallest integral value of $x$ such that $2x^2-2x+1=(2m+1)^2$ and $x \ge 10^{15}$.
| According to Stefan4024, the solution for the Pythagorean triples are well-known:
$$\begin{pmatrix} x-1 \\ x \\ 2m+1 \end{pmatrix}=
\begin{pmatrix}
\frac{(\sqrt{2}+1)^{2k+1}-(\sqrt{2}-1)^{2k+1}}{4}-\frac{1}{2} \\
\frac{(\sqrt{2}+1)^{2k+1}-(\sqrt{2}-1)^{2k+1}}{4}+\frac{1}{2} \\
\frac{(\sqrt{2}+1)^{2k+1}+(\sqrt{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding limit of sequence, done right: $a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$? I need to find the limit of this sequence:
$a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$
So I multiply with this since $(a-b)(a+b)=(a^2-b^2)$
$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$
And get $\dfrac{9-n+9}{\sqrt{n^2+9}+\sqrt{n^2-n... | After multiplying the initial equation by $$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$$
You should get $(n^2+9) - (n^2 - n +9) = n$ in the numerator, i.e., you should have this:
$$\frac{n}{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}$$
Now try dividing numerator and denominator by $n$:
$$\lim_{n\to \infty} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $ Given the followning limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $$
To find limit I have made following steps:
*
*Let $ x = y $ ,then limit equals $0$
*Let $ x... | Hint,
Assume $x = r \cos \theta$ and $y = r \sin \theta, \theta = constant$
Then, the limit changes to
$$ \lim \limits_{r \rightarrow \infty}\left ( \dfrac {r^2 \sin \theta \cos \theta}{r^2}\right)^{r^2 \cos^2 \theta} \\ \Rightarrow \lim \limits_{r \rightarrow \infty} (\sin \theta\cos \theta)^{r^2 \cos^2\theta} = 0$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Quadric obtained by rotation of a line I have a line of equations:
\begin{align}r:
\begin{cases}
x = 2+2t \\
y = 2 - t \\
z=t
\end{cases}
\end{align}
and another line of equations:
\begin{align}s:
\begin{cases}
x = 3 \\
y = 1 \\
z=k
\end{cases}
\end{align}
I have to determine what quadric I get when i rotate line $r$ a... | We have line of equation
\begin{align}r:
\begin{cases}
x=f(t)=2+2t\\
y=g(t)=2-t\\
z=h(t)=t
\end{cases}
\end{align}
and the axis of the rotation
\begin{align}s:
\begin{cases}
x=x_0+at=3\\
y=y_0+bt=1\\
z=z_0+ct=k
\end{cases}
\end{align}
Let's choose a generic point on line $s$ and let it be $C(3,1,0)$.
We have to find th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative va... | By Weierstrass Substitution:
By symmetry, we can half the integration interval as
$$
\int_0^{2 \pi} \frac{1}{5-4 \cos x} d x=2 \int_0^\pi \frac{1}{5-4 \cos x} d x
$$
Putting $t=\tan \frac{x}{2}$ transforms the integral into
$$
\begin{aligned}
I & =2 \int_0^\pi \frac{1}{5-\frac{4\left(1-t^2\right)}{1+t^2}} \cdot \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
| Given equation : $(x^2-3x+1)^2=4x^2-12x+9$ ....(1)
To solve this equation we need the identity,
$a^2-2ab+b^2=(a-b)^2$
on observing the R.H.S. of the equation (1) we can write $4x^2-12x+9$ as
$(2x)^2-2\times2x\times3+3^2$
Now from the above identity, it can be written as $(2x-3)^2$
Thus, equation (1) becomes $(x^2-3x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Simplifying a radical with complex fractions So I understand to simplify this:
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}}
$$
I can just multiply
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}} \cdot\frac{2t^4}{2t^4}
$$
and get
$$
\frac{-3}{t\sqrt{\frac{1}{4t^6} - 1}}
$$
B... | One may write
$$
\begin{align}
\frac{\frac{-3}{2t^4}}{\left|\frac{1}{2t^3}\right|\sqrt{\frac{1}{4t^6} - 1}}&=\frac{\frac{-3}{2t^4}}{\left|\frac{1}{2t^3}\right|\sqrt{\frac{1-4t^6}{4t^6}}}
\\&=\frac{\frac{-3}{2t^4}}{\left|\frac{1}{2t^3}\right|\frac{\sqrt{1-4t^6}}{\sqrt{4t^6}}}
\\&=\frac{\frac{-3}{2t^4}}{\frac{1}{2\left|t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$... | $\begin{vmatrix}
x^2+y^2&x&y&1\\
5^2+7^2&5&7&1\\
6^2+6^2&6&6&1\\
2^2+(-2)^2&2&-2&1
\end{vmatrix}=0$
$-12(x^2+y^2)+48x+72y+144=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to integrate this integral?. How to integrate $$\int\limits_{1}^{3}\cos(5x^2)\,\mathrm dx \, ?$$ Doesn't seem to work by using intergation by parts or substitution.
| We can simply calculate the indefinite integral:
Substitute $u=\sqrt{\frac{10}{\pi}}x$ to get a Fresnel integral:
$$\int \cos (5x^2)\,\mathrm d x=\sqrt\frac{\pi}{10}\int\cos \left( \frac{\pi}{2}u^2\right)\,\mathrm d u=\sqrt\frac{\pi}{10} C(u)=\sqrt\frac{\pi}{10}\,C\left( \sqrt\frac{10}{\pi} x\right) +C$$
Now use the fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any succ... | Switch the sign of columns 2. and 4., then multiply column 2. and 3. by 3 and finally add them up to get zero in every row:
Col.1 - 3 Col.2 + 3 Col.3 - Col.4 = 0
So columns are linearly dependent, hence the determinant is zero for any $a\ldots d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 5
} |
Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$
I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
| Multiply both the numerator and the denominator by $x(1-x)(1+x)$ and then use the fact that $(1-x)(1+x)=1-x^2$ to see that
\begin{align}
\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}&=\left(\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}\right)\frac{x(1-x)(1+x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Show that a $f(r\cos\theta, r\sin\theta)=rh(\theta)$ is continuous on $\mathbb{R}^2$. "Vector Calculus ..." - Hubbard & Hubbard: Exercise 1.20 There's this exercise in Hubbard's book:
Let $ h:\Bbb R \to \Bbb R $ be a $C^1$ function, periodic of period $2\pi$, and define the function $ f:\Bbb R^2 \to \Bbb R $ by
$$f... | First consider the limit of $f$ as you approach some point $\begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}$ for which $s>0$.
\begin{align*}
\lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}}f\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nonlinear equations in 3 variables Solve for $x,y\ \mbox{and}\ z$:
$$
\left\{\begin{array}{rcr}
x + y + z & = & 2
\\[1mm]
\left(x + y\right)\left(y + z\right) + \left(y+z\right)\left(z+x\right) +
\left(z + x\right)\left(x + y\right) & = & 1
\\[1mm]
x^{2}\left(y + z\right) + y^{2}\left(x + z\right) + z^{2}\left(x ... | Way:
$1)$ express LHS-s via elementary symmetric polynomials;
$2)$ consider related cubic equation.
$1)$ Elementary symmetric polynomials for this problem are:
$e_1(x,y,z) = x+y+z$;
$e_2(x,y,z) = xy+xz+yz$;
$e_3(x,y,z) = xyz$.
So,
$$x+y+z= e_1;$$
$$(x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y)\\ =3(xy+yz+xz)+x^2+y^2+z^2 \\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Angle of a triangle using trigonometry. If in a triangle ABC , cos A+ cos C= sin B then what is the measurement of angle A? I have tried to solve it using trigonometric identities but failed to solve it.
| $$\cos A + \cos C = \sin(180-A-C)$$
$$\cos A + \cos C = \sin(A+C)$$
$$\cos A + \cos C = \cos A \sin C + \cos C \sin A$$
$$\cos A (1-\sin C) + \cos C (1-\sin A) = 0$$
If $\sin A = 1$ or $\sin C= 1$, we see that this will hold automatically. Otherwise, we have that
$$\frac{\cos A}{1-\sin A} + \frac{\cos C}{1-\sin C} = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the coefficient of $x^{100}$ in the power series representing the function $\dots$ Find the coefficient of $x^{100}$ in the power series representing the function:
$$f(x)=(x+x^2+x^3+ \cdots) \cdot (x^2+x^3+x^4 \cdots) \cdot (x^3+x^4+x^5 \cdots)$$
Here is what I have so far:
$x+x^2+x^3+ \cdots = x(1+x+x^2+x^3+ \cdo... | As others have noted in the comments, you should have
$$F(x)=\frac{x^6}{(1-x)^3}=x^6\sum_{n\ge 0}\binom{n+2}2x^n=\sum_{n\ge 6}\binom{n-4}2x^n\;.$$
The general rule here is
$$\frac1{(1-x)^{m+1}}=\sum_{n\ge 0}\binom{m+n}mx^n\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When will a parametric solution generate all possible solutions? I was looking for integer solutions to this equation:
$$a^2+b^2+c^2=3d^2$$
And found a parametric solution. Given r, s, t:
$$a=r^2+s^2-t^2+2t(r+s)$$
$$b=r^2-s^2+t^2+2s(t-r)$$
$$c=-r^2+s^2+t^2+2r(t-s)$$
$$d=r^2+s^2+t^2$$
My question is will this generate ... | You almost got it. The complete rational parameterization to
$$a^2+b^2+c^2=3\tag1$$
is given by,
$$a=\frac{r^2+s^2-t^2-2t(r+s)}{r^2+s^2+t^2}\\b=\frac{r^2-s^2+t^2-2s(r+t)}{r^2+s^2+t^2}\\c=\frac{-r^2+s^2+t^2-2r(s+t)}{r^2+s^2+t^2}\tag2$$
Proof: For any solution $a,b,c$, one can find $r,s,t$ using the simple formulas,
$$r=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction
$$ 21 \mid (2^{4n}+5) $$
So I assume:
$ 2^{4k}+5= 21p$
to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ =
$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 1... | The statement is true for $n=1$. Suppose it holds for $k$: $2^{4k}+5=21p$, so $2^{4k}=21p-5$, so
$$
2^{4(k+1)}+5=2^{4k}\cdot 16+5=
16(21p-5)+5=21\cdot 16p-80+5=21\cdot16p-75
$$
Ops! Something seems wrong. We have proved that
if $21\mid 2^{4k}+5$, then $21\nmid 2^{4(k+1)}+5$
Actually, $3\mid 2^{4n}+5$ (prove it), but ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simple way to calculate $n! \pmod p$ I have the exercise "Calculate $10! \pmod{13}$".
I have the following two approaches to solve the exercise:
*
*Brute force approach
$$
1! \equiv 1 \pmod{13} \\
2! = 2 \cdot 1! \equiv 2 \cdot 1 \equiv 2 \pmod{13} \\
3! = 3 \cdot 2! \equiv 3 \cdot 2 \equiv 6 \pmod{13} \\
\cdots \... | Semi brute force. $14\equiv 1 \mod 13$ so toss out $2,7$. $27 \equiv 1$ so toss out $3,9$. $40 \equiv 1$ so toss out $4,10$ and $8,5$. What do we have left.
Just $6$. $10! \equiv 6\mod 13$
because $10! =(2*7)(3*9)(4*10)6 (5*8)\equiv 6 \mod 13$
Hmm, that's not very satisfying, is it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Show $\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$ I want to verify the following inequality:
Let $a, b$ be non negative number
$$\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$$
I decided to analyse the sign of $$\frac{a^2}{b^2} + \frac{b^2}{a^2} +... | Let $t=a/b+b/a$. Then this becomes $t^2+1\geq 2t$, which is same as $(t-1)^2 \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... | As is clear, this question is invariant under horizontal displacement i.e. you can replace $f(x)=(x-\dfrac{a}{2})^2+1-\dfrac{a^2}{4}$ with $f(x)=x^2+1-\dfrac{a^2}{4}$. Here after we use the latter notation. Define $b=1-\dfrac{a^2}{4}$therefore $$f(x)=x^2+b$$assume such a $x_0$ exists. We cannot have both $x_0$ and $x_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 6
} |
Precalc coursework - Is this an error? I am enrolled in a precalc math course at a university studying online. I haven't studied math in over 35 years so I'm a bit rusty. Why would my online math coursework have errors? I end up torn between feeling like I must just not understand something and certainty that there is ... | You are right. There is a missing $x.$ The final result should be $\dfrac{x+4}{2x^2}.$ Note that
$$\begin{align}\frac{1}{x+3}\left(\frac{x+7}{2x}+\frac{6}{x^2}\right)& = \frac{1}{x+3}\left(\frac{(x+7)\cdot x}{2x\cdot x}+\frac{6\cdot 2}{x^2\cdot 2}\right)\\ &= \frac{1}{x+3}\left(\frac{x^2+7 x}{2x^2}+\frac{12}{2x^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$
I think $uvw$ does not help here.
My another similar inequality is very easy:
with the same condition prove tha... | Another solution using KKT conditions (or Lagrange multiplier)
Let
\begin{align}
L = \ln \frac{a+b+c}{3} - \frac{1}{53}\ln \frac{a^4+b^4+c^4}{3}
- t ((a+b)(b+c)(c+a) - 8).
\end{align}
From $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial b} = \frac{\partial L}{\partial c} = 0$, we have
\begin{align}
0 = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Algebraic expansion of $(1+x)^{1/x}$ is? i have to solve
$ \lim_{x\to 0} \frac{{(1+x})^{1/x}-e+ex/2}{x^2}$ and in the hint it is advised to use the algebraic expansion of $(1+x)^{1/x}$.
Also i know that the algebraic expansion of $(1+x)^{m} = 1+mx+m(m-1){x^2}/2+...$ but this is in the case where m is an integer.
So pl... | Note that you can develop
$$
\begin{gathered}
\left( {1 + x} \right)^{\,1/x} = \exp \left( {\frac{1}
{x}\ln \left( {1 + x} \right)} \right) = \exp \left( {1 - \frac{x}
{2} + \frac{{x^{\,2} }}
{3} + O\left( {x^{\,3} } \right)} \right) = \hfill \\
= \exp \left( 1 \right)\exp \left( { - \frac{x}
{2}} \right)\exp \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Integration of $\frac{2x^2}{\sqrt{x^2-1}}\, dx$ Integration of $\dfrac{2x^2}{\sqrt{x^2-1}}\,\mathrm dx$
My try:
Let $u=\sqrt{x^2-1}$
Then $x=\sqrt{u^2+1}$
$x \,\mathrm dx =u\,\mathrm du$
$$\int \frac{2u(u^2+1)}{u\sqrt{u^2+1}} \, \mathrm du$$
$$=2\int \sqrt {u^2+1} \, \mathrm du$$
True ? and what about the last integra... | my hint:
$$\int \frac{2x^2}{\sqrt{x^2-1}}\ dx=\int \frac{2(x^2-1)+2}{\sqrt{x^2-1}}\ dx=2\int\sqrt{x^2-1}\ dx+2\int\frac{1}{\sqrt{x^2-1}}\ dx$$
both integrals can be evaluated by substituting $u=\sec \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Arithmetic Mean/Geometrix Mean Inequality of Degree 3 For $a,b$ and $c\geq0$;
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
Is there a simple way to deduce this result from the degree 2 verson?
That is:
$$\frac{a+b}{2}\geq\sqrt{ab}$$
Some sort of substitution involving $a,b,c \ $ for one of the variables was what I had in min... | I think, the previous method is the best (with using AM-GM for two variables),
but we can make also the following proof by the same AM-GM.
Since for non-negatives $x$, $y$ and $z$ by AM-GM we have
$\frac{x^2+y^2}{2}\geq xy$ and $\frac{x^2+z^2}{2}\geq xz$ and $\frac{y^2+z^2}{2}\geq yz$,
after summing we obtain $x^2+y^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
The matrix $e^A$ is defined by $e^A=\Sigma_{k=0}^{\infty}\frac {A^k}{k!}$ Suppose M=$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. Calculate $e^M$ The matrix $e^A$ is defined by $e^A=\Sigma_{k=0}^{\infty}\frac {A^k}{k!}$ Suppose M=$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. Calculate $e^M$.
I did some calculating with rea... | Observe
\begin{align}
M=
\begin{pmatrix}
1& 0\\
0 & 1
\end{pmatrix}
+
\begin{pmatrix}
0& 1\\
0 & 0
\end{pmatrix}=: I+ X
\end{align}
where $[X, I] = 0$ (of course since identity commutes with anything). Hence it follows
\begin{align}
\exp[M] = \exp[I]\exp[X].
\end{align}
Next, observe
\begin{align}
X^2 = 0 \ \ \Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Summation Proof prove that:
I am having trouble coming up with a rigorous proof for this.Can I use induction? Can anyone demonstrate?
$$\sum_{i={2^n+1}}^{{2^{n+1}}}{1/i}\ge 1/2$$
| $$
\sum_{i = 2^n + 1}^{2^{n+1}} \frac{1}{i} = \frac{1}{2^n + 1} + \frac{1}{2^n + } + \cdots + \frac{1}{2^n + 2^n}
$$
$$
> \frac{1}{2^n + 2^n} + \frac{1}{2^n + 2^n} + \cdots + \frac{1}{2^n + 2^n} = \frac{2^n}{2^n + 2^n} = \frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit involving nested roots How do I solve it? Is there some crafty algebraic step I need to follow?
I'm guessing it's -∞ in the end..
$$ \lim_{x\to-∞} \sqrt{x^2-\sqrt{x^2+1}}+x $$
| Rewrite it as follows: for $x< 0$,
$$\begin{align}
\sqrt{x^2-\sqrt{x^2+1}}+x &= \sqrt{x^2-|x|\sqrt{1+\frac{1}{x^2}}}+x = |x|\sqrt{1-\frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}}+x \\
&= -x\left(1-\sqrt{1-\frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}}\right)
\end{align}$$
Writing $u(x)\stackrel{\rm def}{=} \frac{1}{|x|}\sqrt{1+\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $ Let $x,y,z > 0$ and $xyz=8.$ Prove that
$$
\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12
$$
I have tried with AM–GM inequality but no result.
| We'll prove that $\sum\limits_{cyc}\frac{(x+y-1)^2}{z}\geq\frac{27}{2}$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $w=2$ and we need to prove that $f(v^2)\geq0$, where
$f(v^2)=9u^2v^2-5uw^3-3uv^2w^3-\frac{5}{4}w^4+\frac{1}{4}v^2w^2$.
But $f$ is a linear function, which says that it's enough to prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating the improper integral $\int_0^\infty \frac{x\cos x-\sin x}{x^3} \cos(\frac{x}{2}) \mathrm dx $ I've been working through the following integral and am stumped:
$$\int_0^\infty \frac{x\cos x-\sin x}{x^3}\cos\left(\frac{x}{2}\right)\mathrm dx$$
Given the questions in my class that have proceeded and followed t... | Inspired by Felix Marin's calculation using integration by parts.
Observe
\begin{align}
\int^\infty_0 \frac{x\cos x-\sin x}{x^3}\cos\frac{x}{2}\ dx=&\ \frac{1}{2}\int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^3}e^{ix/2}\ dx\\
=&\ \frac{-1}{4} \int^\infty_{-\infty} [x\cos x-\sin x] e^{ix/2}\ d\left(\frac{1}{x^2} \right).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
} |
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking ... | An elementary approach
\begin{align}
I &= \int_0^1 \frac{\arcsin x\arcsin\frac x{\sqrt2}}{\sqrt{2-x^2}} \, dx \\
&= \frac12 \int_0^1 \arcsin x \ d\left(\arcsin^2\frac x{\sqrt2} -\frac{\pi^2}{16}\right)\\
&\overset{ibp}= \frac12 \int_0^1 \left(\frac{\pi^2}{16}-\arcsin^2\frac x{\sqrt2} \right)\frac1{\sqrt{1-x^2}}dx
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find all k such that $\sigma (k)=165$ where $\sigma$ is the sum of divisors $165=(3) (5) (11)$
$\sigma (p^a)$=$p^{a+1}-1 \over {p-1}$=$3$
$p^{a+1}-1$=$3p-3$
$p^{a+1}=3p-2$
Got stuck here. How do I proceed?
| Since it seems that you are looking for the "hard" way:
Let $n$ be a positive integer such that $\sigma(n)=165.$ Then $\prod_{p\mid n}(p^{\alpha_p+1}-1)/(p-1)=3\cdot5\cdot11,$ where the product is taken over all primes dividing $n$ and where $\alpha_p$ is the exponent of $p$ in the prime decomposition of $n.$ Now note ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to solve a particular well ordering problem I want to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$ with well ordering principle.
I know that $$n \in N $$
because the numbers 1,2,3... are discrete and are positive. I was thinking of using a set of na... | Yes, you can make it like that. It is not different than induction, though.
Let $E$ be the set of $n\in\mathbb N$ such that your equality does not hold. Note that $1\not\in E$. Let $k\geq2$ be the minimum of $E$ (here is where one uses the principle). Then $k-1\not\in E$, so it satisfies the formula:
$$
\frac{1}{1\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
| Yes, $x+1$ is a factor of $x^{2016}-1$ and
$$x^{2016}-1=(x+1)(x^{2015}-x^{2014}+x^{2013}-\cdots-x^2+x-1)$$
Then
$$G(x)=1+x^2+x^4+\dots+x^{2014}$$
More generally, $x-r$ is a factor of $P(x)$ if and only if $P(r)=0$. In this case, $P(x)=x^{2016}-1$ and $P(-1)=0$, so $x+1$ is a factor of $x^{2016}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the average rate of change of $S(x) = -2x^2 + 14x - 12$ I have to show that the average rate of change of $S(x) = -2x^2 + 14x - 12$ in the interval $[x,x+h]$ is $-4x - 2h + 14$ and so far I did:
$$A(x) = \frac{S(x+h)-S(x)}{(x+h)-(x)} = \frac{(-2(x+h)^2+14(x+h)-12)-(-2x^2+14x-12)}{(x+h)-x} = \frac{-2(x+h)^2+14(... | Starting from
$$\frac{-2(x+h)^2+14(x+h)+2x^2-14x}h$$
expand all terms in the numerator and simplify:
$$=\frac{-2(x^2+2hx+h^2)+14x+14h+2x^2-14x}h$$
$$=\frac{-2x^2-4hx-2h^2+14h+2x^2}h$$
$$=\frac{-4hx-2h^2+14h}h$$
$$=-4x-2h+14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a convergence for the series $ \sum_{i=0}^{\infty} \frac{(x-i)^i}{i!} $ The following series converges to exponential.
$\sum_{i=0}^{\infty} \frac{x^i}{i!} = e^{x}$
Is the convergence of the following series known?
$\sum_{i=0}^{\infty} \frac{(x-i)^i}{i!}$
| $$a_n=\frac{(x-n)^n}{n!}\implies A=\frac{a_{n+1}}{a_n}=\frac1{n+1 }\frac{ (x-n-1)^{n+1}} {(x-n)^n }=\frac{-1}{n+1 }\frac{ (n+1-x)^{n+1}} {(n-x)^n }$$ $$B=\frac{ (n+1-x)^{n+1}} {(n-x)^n }\implies\log(B)=(n+1)\log(n+1-x)-n\log(n-x)$$ Now, for large values of $n$, use Taylor$$p \log (p-x)=p \log \left(p\right)-x-\frac{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+... | Using $$\sin (a+b+c)-\sin a-\sin b-\sin c $$
$$= 2\cos\left(\frac{2a+b+c}{2}\right)\sin \left(\frac{b+c}{2}\right)-2\sin \left(\frac{b+c}{2}\right)\cos \left(\frac{b-c}{2}\right)$$
So $$ = 2\sin \left(\frac{b+c}{2}\right)\left[\cos \left( \frac{2a+b+c}{2}\right)-\cos \left(\frac{b-c}{2}\right)\right]$$
$$ = -4\sin \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving Recurrence Relation T(n) = T(n-1) + n + 2 I am getting stuck solving this recurrence relation.
T(1) = 1
T(n) = T(n-1) + n + 2
I continued to expand it:
$T(n) = T(n-1) + n + 2 = [T(n-2) + (n-1) + n + 2] + 2$
$= T(n-2) + (n-1) + n +4 = [T(n-3) + (n-2) + (n-1) + n + 2] + 4$
$= T(n-3) + (n-2) + (n-1) + n + 6$
$= T(... | HINT:
$$T(n) - T(n-1) = n +2 = T(n-1) - T(n-2) + 1 \implies T(n) - 2T(n-1) + T(n-2) = 1$$
Now in the same manner:
$$T(n) - 2T(n-1) + T(n-2) = 1 = T(n-1) - 2T(n-2) + T(n-3) \implies $$
$$T(n) - 3T(n-2) + 3T(n-2) - T(n-3) = 0$$
Now this is a homogenous recursive relation, which can be easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Write the given Determinant as the product of two Determinants
Question Statement:-
Prove that:-
$$\begin{vmatrix}
b+c-a-d & bc-ad & bc(a+d)-ad(b+c)\\
c+a-b-d & ca-bd & ca(b+d)-bd(c+a)\\
a+b-c-d & ab-cd & ab(c+d)-cd(a+b)\\
\end{vmatrix}\qquad\qquad=2(a-b)(b-c)(c-a)(a-d)(b-d)(c-d)$$
Attempt at a solution:-
I tried ... | I don't know how to do it your way, but here is another one. It goes by row-reduction.
In a first step, we perform row reduction in the following way: replace $R_1$ with $R_1-R_2$ and $R_3$ with $R_3-R_1$. And we want to replace $R_2$ with $R_1+R_2$. This can be obtained as $$2\left(R_2+\frac{R_1-R_2}2\right)=R_1+R_2.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is there a general formula for $I(m,n)$? Consider the integral
$$I(m,n):=\int_0^{\infty} \frac{x^m}{x^n+1}\,\mathrm dx$$
For $m=0$, a general formula is $$I(0,n)=\frac{\frac{\pi}{n}}{\sin\left(\frac{\pi}{n}\right)}$$
Some other values are $$I(1,3)=\frac{2\pi}{3\sqrt{3}}$$ $$I(1,4)=\frac{\pi}{4}$$ $$I(2,4)=\frac{\pi}{2\... | We shall compute it in two steps. First, perform the substitution $y = x^n$ in order to get
$$I(m,n) = \int \limits _0 ^\infty \frac {y ^{\frac m n}} {1 + y} \frac 1 n y ^{\frac 1 n - 1} \ \Bbb d y = \frac 1 n \int \limits _0 ^\infty \frac {y ^{\frac {m+1} n - 1}} {1 + y} \ \Bbb d y .$$
Now perform the change $t = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \... | HINT: You know what $\sin^2\alpha+\cos^2\alpha$ is, so the problem boils down to sorting out $\sin\alpha\cos\alpha$. But
$$(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\;,$$
and you can solve this for $\sin\alpha\cos\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
A problem related to arithmetico-geometric sequence Question:
Find the sum of the series: $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ... = 1+\sum_{n=1}^\infty \frac{(4n-2)}{3^n}$
My doubt:
I have taken $\frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ...$ to be in A.G.P. and ... | It's not a standard AGP. An AGP has the recurrence relation $u_{n+1}=r*u_n+b$, but here that isn't the case. So that's why you can't plug it into the formula for an AGP!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
functional equation of type $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$
If $f:\mathbb{R}-\{-1\}\rightarrow \mathbb{R}$ and $f$ is a differentiable function that satisfies $$f(x+f(y)+xf(y)) = y+f(x)+yf(x)\forall x,y \in \mathbb{R}-\{-1\}\;,$$ Then value of $\displaystyle 2016(1+f(2015)) = $
$\bf{My\; Try::}$ Using partial Differ... | (Work in progress; I was unexpectedly called away. The aim of this post is to do it without assuming differentiability.)
Roots
To find the roots of $f$: suppose $f(y) = 0$.
Then $f(x) = y+f(x)+y f(x)$ and so $(1+f(x))y = 0$; therefore $f(x) = -1$ for all $x$, or $y=0$.
So we have two possibilities: $f(x) = -1$ for all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Deriving the equation of a circle I have two points, $A=(-1,3)$ and $B=(2,7)$. There is a third point $P=(x,y)$.
I have found $|AP|$ and $|BP|$ in terms of $x$ and $y$. These are:
$|AP| = \sqrt{(-1-x)^2 +(3-y)^2}$
$|BP| = \sqrt{(2-x)^2 +(7-y)^2}$
I have shown that the set $S = \{P:|AP| = |BP|\}$ is a straight line and ... | $$|AP|^2=(x+1)^2+(y-3)^2$$
$$|BP|^2=(x-2)^2+(y-7)^2$$
The equation of $C$ may be worked out as follows:
$$|AP|=\alpha|BP|$$
$$|AP|^2=\alpha^2|BP|^2$$
$$(x+1)^2+(y-3)^2=\alpha^2((x-2)^2+(y-7)^2)$$
$$x^2+2x+1+y^2-6y+9=\alpha^2(x^2-4x+4+y^2-14y+49)$$
$$(\alpha^2-1)x^2+(\alpha^2-1)y^2-(\alpha^2+\tfrac12)(4x)-(\alpha^2-\tfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
... | $$17x\equiv 1\pmod{41}\\-24x\equiv -40\pmod{41}\\3x\equiv 5\pmod{41}\\3x\equiv-36\pmod{41}\\x\equiv -12\equiv29\pmod{41}$$
Multiplying $x\equiv29\pmod{41}$ by $14$ also gives a correct solution though I'm not quite sure if it's a coincidence or not.$$\\17x\equiv14\pmod{41}\\-24x\equiv14\pmod{41}\\-12x\equiv 7\pmod{41}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
Explain this equality: $-90^\circ + \tan^{-1}\frac{X_C}{R} = - \tan^{-1}\frac{R}{X_C}$
$$-90^\circ + \tan^{-1}\frac{X_C}{R} \;=\; - \tan^{-1}\frac{R}{X_C}$$
Please explain how both are equal.
| $$\text{Let} \ \ \arctan \frac{1}{x} = \theta$$
$$\frac{1}{x} = \tan\theta$$
$$x = \cot \theta = \tan \left(\frac{\pi}{2} - \theta\right)$$
$$\arctan x = \frac{\pi}{2} - \theta$$
So, $$ \arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
points on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is minimum
Find point on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is, minimum
$\bf{My\; Try::}$ Let $(x,y)$ be any point on the curve $x^2+2y^2=6\;,$ Then we have to
minimize $\displaystyle \left|\frac{x+y-7}{\sqrt{2}}\right|... | Let $t = x+y$, you've shown $-3 \le t \le 3$, thus the function $f(t) =\dfrac{ (t-7)^2}{2}$ has a graph a parabola with the axis of symmetry at $x = 7$, and it is decreasing on $[-3,3]$, thus the min is $f(3) = 8$. Thus the min distance is $2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to symbolically solve a system of linear equations A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \fra... | One can also make use of the Mathematica programming language and proceed as follows:
where the lines with MatrixForm[$\cdots$] were just to check if the matrices were fine.
The command LinearSolve[matrix, result] finds the solution and the added // MatrixForm is just for pretty printing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that$\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$=$\frac{1-{x^2}}{2x}$ Well, I was trying to find $\tan40$ in terms of $\tan25$=x.
So, I expanded 40 as 25+15 and got the value of $\tan40$ in terms of $x$ as $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$.
Now, I solved $\tan40$ alternatively as
$$\frac{\tan155-\tan115}{1+\tan15... | When $\frac{1-x^2}{2x}$ is equated to $\frac{x+2−\sqrt{3}}{1+(\sqrt{3}-2)x}$, it gives rise to: $$x^3-3(\sqrt{3}+2)x^2-3x+(\sqrt{3}+2)=0$$
Wolfram gives three roots:$$x≈-0.700208,\,\,\,x≈0.466308,\,\,\,x≈11.4301$$
Also, note that, $\tan{25}≈0.4663076...$
It is not a satisfying "exact" proof, but something to get a sens... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove by induction, that $ \sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!}$ If I'm not wrong,
$$\sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!},$$
but I am having trouble proving it by induction...
If $n=1$ the formulas coincide.
Sup $n=k$ is valid: $$\sum_{i=1}^k \frac{i}{(i+1)!}= \frac{1}{(k+1)(k-1)!}.$$
Th... | Try this in your inductive step:
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+2)!} $$
Notice the denominator is $(k + 2)!$ instead of $(k+1)!$, which makes simplification easier:
$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+2)!} $$
$$ = \frac{(k + 2)k}{(k+2)!} + \frac{k+1}{(k+2)!} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Laplace transform of a differential equation: $y'+y=e^{-3t}\cos(2t)$ Given $y(0)=0$, solve
$$y'+y=e^{3t}cos(2t) $$
Steps:
$$sY(s)-y(0)+Y(s)=\frac{s+3}{(s+3)^2+2^2}$$
$$Y(s)(s+1)=\frac{s+3}{(s+3)^2+2^2}$$
$$Y(s)=\frac{s+1}{(s+1)((s+3)^2+2^2)}+\frac{2}{(s+1)((s+3)^2+2^2)}$$
$$Y(s)=\frac{1}{(s+3)^2+2^2}+\frac{\textbf{2... | Hint. You may use a partial fraction decomposition to get
$$
\frac{2}{(s+1)((s+3)^2+2^2)}=\frac{1}{4(s+1)}-\frac14\cdot\frac{(s+3)}{(s+3)^2+2^2}-\frac12\cdot\frac{1}{(s+3)^2+2^2}
$$ then each term can be classically reversed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Dice problem for the n space There is a 6-faced dice and every time you roll it, you move forward the same number of spaces that you rolled. If the finishing point is “n” spaces away from the starting point, how many possible ways are there to arrive exactly at the finishing point?
| Your questions deals with the number of ["compositions"](https://en.wikipedia.org/wiki/Composition_(combinatorics) of an integer.
Although there is almost surely no closed form formula for it, here is a general technique for obtaining results for a given $n$ by using generating functions.
I will present it through an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to direc... | $A:=\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\enspace\enspace\enspace$ Shift: $\enspace C_2\to C_1$, $C_3\to C_2$, $C_1\to C_3\enspace$ => $\enspace B$
$B:=\begin{vmatrix}
b^2+bc & c^2+bc & -bc\\
-ac & c^2+ac & a^2+ac \\
b^2+ab & -ab & a^2+ab
\end{vmatrix}$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y ... | My another solution inspired by Gordon's solution.
Suppose that $x \leq y \leq z$ Then $\sqrt{x} \leq \sqrt{y} \leq \sqrt{z}$ and
$$
\frac{1}{\sqrt{x}} \geq \frac{1}{\sqrt{y}} \geq \frac{1}{\sqrt{z}}.
$$
Now by rearrangement inequality we get
$$
1 \geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{1}{\sqrt{xy}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How does $ab+ac+bc =0$ imply that $a+b+c = \pm 1$? Show that $$ A=
\begin{bmatrix}
a & c & b \\
b & a & c \\
c & b & a \\
\end{bmatrix}
$$ is orthogonal if and only if $a^2+b^2+c^2=1$ and $a+b+c= \pm 1$.
For a matrix to be orthogonal it must satisfy $A^T A^{-1} = I$ where $I$ is ... | The comments should be enough,
but
if $a^2+b^2+c^2 = 1$
and
$ab+ac+bc = 0$,
then
$(a+b+c)^2
=a^2+b^2+c^2+2(ab+ac+bc)
=1
$
so
$a+b+c = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what kind of test should I use to determine if this sequence converge or diverge? what kind of test should I use to determine if this sequence converge or diverge?
$$\sum_{n=1}^\infty \left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right)$$
| By comparison test $$sin\frac{1}{n}-sin \frac{1}{n+1}=\\2sin(\frac{1}{n}-\frac{1}{n+1})cos(\frac{1}{n}+\frac{1}{n+1})=\\2sin(\frac{1}{n(n+1)})cos(\frac{2n+1}{n(n+1)})\\\leq2sin(\frac{1}{n(n+1)}) \\2\frac{1}{n(n+1)} \leq2 \frac{1}{n^2}\\$$then
$$\sum sin\frac{1}{n}-sin \frac{1}{n+1} \leq \sum \frac{2}{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now... | X(X+1)(X+3)
Therefore (X+1)(X+3) = X^2+4X+3.
Therefore multiplying this by X you get X^3+4X^2+3X.
So yes well done that is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
finding the limit of an expression with roots I need to find the limit of the expression:
$$
\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}
$$
It is given that $b>2$. In the class we were encouraged to use the method of multiplying both the numerator and the denominator with the numerator when the limit can't be ... | Multiply by $\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$
$$\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$$
$$\lim_{x\to b}\frac{b-x}{(x-b)(x+b)(\sqrt{b-2} + \sqrt{x-2})}=\frac{-1}{(b+b)(\sqrt{b-2}+\sqrt{b-2})}=\frac{-1}{4b\sqrt{b-2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side?
| You may note that $3+2\sqrt{2}=(1+\sqrt{2})^2$. Hence, we have
$$
\sqrt{2}\sqrt{3+2\sqrt{2}}-1=\sqrt{2}(1+\sqrt{2})-1=\sqrt{2}+2-1=\sqrt{2}+1,
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equi... | Another solution:
Prove: $$\left ( a+ b- c \right )^{a}\left ( b+ c- a \right )^{b}\left ( c+ a- b \right )^{c}\leq a^{a}b^{b}c^{c}$$
By weight power mean:
$$\sqrt[a+ b+ c]{\left ( \frac{a+ b- c}{a} \right )^{a}\left ( \frac{b+ c- a}{b} \right )^{b}\left ( \frac{c+ a- b}{c} \right )^{c}}\leq \frac{1}{a+ b+ c}\left ( a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
UK 1998, Show that $hxyz$ is a perfect square. Problem: [UK 1998] Let $x,y$ and $z$ be positive integers such that $$\frac{1}{x}-\frac{1}{y}=\frac{1}{z}.$$ Let $h=(x,y,z).$ Prove that $hxyz$ is a perfect square.
My Attempt: Of course $y>x$ and therefore let $y=x+a$ for some $a>0.$ Then $$z=\frac{x(x+a)}{a}.$$ Also $$h... | Let's try your argument with $x = 15; y = 40; z = 24$ where $\frac 1{15} - \frac 1{40}= \frac{8}{8*15} - \frac{3}{3*40} = \frac 5{5*24} = \frac 1 {24}$ and $h = (15,40,24) = 1$ and $hxyz = 1*3*5*8*5 *2*8 = (3*5*8)^2$.
You argue $40 > 15$ and $a = 40 -15 = 25$.
So $z = 24 = \frac {15*(15 + 25)}{25}$ which is indeed tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \fr... | $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
$$=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x+1)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x^2+2x+1)}{(x+1)(\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is there a closed form of the sum $\sum _{n=1}^x\lfloor n \sqrt{2}\rfloor$ I need to find the n-th partial sum of:
$$\sum _{n=1}^x\lfloor n \sqrt{2}\rfloor$$
Or this sum of a Beatty sequence.
I tried to expand as the following:
$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x\sum _{k=0}^... | Since $\lfloor \sqrt 2 k \rfloor$ and $\lfloor (2+\sqrt 2) k \rfloor$ are complementary Beatty sequences, they partition the integers and we can find a recurrence relation for their sum. To sum the sequence $\lfloor \sqrt 2 k \rfloor$ we can use a well-known formula to add the consecutive integers
$$
\begin{align}
\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
For $y = 2f\left(0.5x+4\right)-3$, derive the equation of: $ f(x)=2x^2-3x+4. $
For $y = 2f\left(0.5x+4\right)-3$,
derive the equation of:
$$
f(x)=2x^2-3x+4.
$$
I don't really understand what this question is asking for. Any input would be greatly appreciated.
| I think that the (admittedly poorly worded) problem is asking for you to express $y$ in terms of $x$ given that $\displaystyle y=2\ f\left(\frac x 2 + 4\right)-3$ and $f(x)=2x^2-3x+4$.
In that case:
\begin{align}
y&=2\ f\left(\frac x 2 + 4\right) - 3\\
&=2\left(2\left(\frac x 2 + 4\right)^2 - 3 \left(\frac x 2 + 4\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving algebraic operations for this simple problem I cannot wrap my heard around solving this:
$$\mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}=2\mu\left[1-\left(\frac{1}{2}\right)^{n+1}\right]$$
I have done this instead:
\begin{align}
\mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}&=\mu\frac{1}{\frac{1}{2}}-\frac{(\frac{... | thanks guys!
What about this one
\begin{align}
\mu\frac{1-(\frac{1}{4})^{n+1}}{\frac{3}{4}}&=\mu\frac{1}{\frac{3}{4}}-\frac{(\frac{1}{4})^{n}(\frac{1}{4})^{1}}{\frac{3}{4}}\\
\end{align}
I don't know how to proceed from this other than
\begin{align} \frac{4}{3}\mu-\mu(\frac{3}{4})^{n+1} \end{align}
which is obviosly w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On the integral $\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}$ and the plastic constant We have,
$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$$
$$P=\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\cdots}}}$$
with golden ratio $\phi$ and plastic constant $P$. If,
$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=\... | HINT:
Let $\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}=y\implies x+y=y^3\iff x=y^3-y$
$dx=3y^2-1$
$x=1\implies y\approx 1.3247179572447458$
$$\int \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}=\int\dfrac{3y^2-1}y\ dy=\dfrac{3y^2}2-\ln|y|+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Reduction formula for integral of $\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx $ The following reduction formula is taken from http://www.sosmath.com/tables/integral/integ15/integ15.html:
$$\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx = -\frac{\sqrt{ax^2+bx+c}}{cx} - \frac{b}{2c} \int \frac{1}{x \sqrt{ax^2+bx+c}} dx$$
*
*I've ... | One may observe that
$$
\begin{align}
\left(-\frac{\sqrt{ax^2+bx+c}}{cx}\right)'&=\frac{\sqrt{ax^2+bx+c}}{c x^2}-\frac{2ax+b}{2cx\sqrt{ax^2+bx+c}}
\\\\&=\frac{2(ax^2+bx+c)-(2ax+b)x}{2cx^2 \sqrt{ax^2+bx+c}}
\\\\&=\frac{bx+2c}{2cx^2 \sqrt{ax^2+bx+c}}
\\\\&=\frac{b}{2c}\cdot\frac{1}{x \sqrt{ax^2+bx+c}}+\frac{1}{x^2 \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
| What you have looks like
$$x^8+x^6-4x^4+x^2+1$$
Which on rearranging becomes $$(x-1)^2(x+1)^2(x^4+3x^2+1)$$
Since the terms in the product are greater than or equal to zero so the product itself is greater than or equal to zero. i.e
$$x^8+x^6-4x^4+x^2+1\ge0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Finding the roots, with multiplicity, of a polynomial with compound angle formula My question is to find the roots, counted with multiplicity, of the polynomial equation
$16x^5-20x^3+5x-1=0$ using the compound angle formula $\sin\left(5\theta\right)=16\sin^5\theta-20\sin^3\theta+5\sin\theta$
So after substituting $x=\... | As already explained, the solutions are:
$$x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$$
Because,
$$\sin(5\theta)=1 \Rightarrow \theta=\frac{1}{5}\left(\frac{\pi}{2}+2k\pi\right)$$
and setting $k=0,1,2,3,4$ we get all five sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Radius of convergence and convergence sets of $\sum\limits_n\frac{2n+1}{(n-1)^2}x^n$ and $\sum\limits_n(-1)^n(\sqrt{n+1}-\sqrt{n})x^n$ I want to find the radius of convergence of the following series and the set of $x\in \mathbb{R}$ in which the series converge.
*
*$$\sum_{n=2}^{\infty}\frac{2n+1}{(n-1)^2}x^n$$
*... | We can calculate the radius of convergence by the formula too: $R=lim|\frac{a_{n}}{a_{n+1}}|$, where $a_{n}$ is the $n^{\text{th}}$ term of the given series.
In your first problem, $a_n=\frac{2n+1}{(n-1)^2}$ and $a_{n+1}=\frac{2(n+1)+1}{n^2}$. So $\frac{1}{R}=lim\frac{(2n+3)(n-1)^2}{(2n+1)n^2}=lim\frac{(1+\frac{3}{2n})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve for all real values of $x$ and $y$ Solve the system of equations for all real values of $x$ and $y$
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$
I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5... | Let $z=x + iy \in \mathbb{C}$, now for $|z|^2 = x^2+y^2 \not = 0$ have that:
$$\left(5x + \frac{5x}{x^2+y^2}\right) + i\left(5y - \frac{5y}{x^2+y^2}\right) = 12 + 4i$$
$$5(x+iy) + \frac{5(x-iy)}{x^2+y^2} = 12 + 4i$$
$$5(x+iy) + \frac{5}{x+iy} = 12 + 4i$$
$$5z + \frac{5}{z} = 12 + 4i$$
$$5z^2 - (12+4i)z + 5 = 0$$
Solvin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\int \frac{x dx}{x^4 + 6x^2 + 5}$? $$\int \frac{x dx}{x^4 + 6x^2 + 5}$$
How to evaluate this integral ?
| $$\int \frac { xdx }{ x^{ 4 }+6x^{ 2 }+5 } =\frac { 1 }{ 2 } \int { \frac { d{ x }^{ 2 } }{ { \left( { x }^{ 2 }+3 \right) }^{ 2 }-4 } } =\frac { 1 }{ 2 } \left[ \int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+5 \right) } } \right] =\frac { 1 }{ 8 } \left[ \int { \frac { d{ x }^{ 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| Let's look at
the area below the part of the curve
from
$(0, 1)$ to $1, 1)$
and the line $y = x$.
The total area is
8 times this.
The area below
$y = x$
is obviously
$1/2$.
If
$x^4+y^4=x^2+y^2
$,
$y^4-y^2
=x^2-x^4
$
or,
trying to be clever,
$y^4-y^2+1/4
=x^2-x^4-1/4+1/2
$
or
$(y^2-1/2)^2
=1/2-(x^2-1/2)^2
$
or
$y^2
=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find an equation for a tangent line to the curve $x^2- y^2=5$ that passes through the point $(1, 1)$. Find an equation for a tangent line to the curve $$ x^2 - y^2 = 5$$ that passes through the point $(1, 1)$.
I realize that I have to use implicit differentiation $$2x - 2y \frac {dy}{dx} = 0$$
$$\frac {dy}{dx} = \frac... | Here's an approach without (implicit) differentation.
A line passing through $(1,1)$ and with slope $m$ has an equation of the form:
$$y=m(x-1)+1 \iff y = mx-m+1$$
The points of intersection of this line and the hyperbola $x^2-y^2=5$ are the solutions to the following system:
$$\left\{\begin{array}{l}
y = mx-m+1 \\
x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Which would you rather have? Q. Which would you rather have, a piece of an 8-inch pie that's been cut into sixths or a piece of a 10-inch pie that's been cut into eights?
A. This is a problem involving sectors. One-sixth of a pie is $\frac {1}{6}$ of $ 2\pi$ radians. The measure of the central angle is $\frac{1}{6} \... | Area per piece of 8 inch pie = ${1\over 6} \times \pi (4)^2 = 8.38$
Area per piece of 10 inch pie = ${1\over 8} \times \pi (5)^2 = 9.82$
So the we will take the piece of 10 inch pie. $\ddot \smile$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinants question with factorials $$A= \begin{pmatrix}
n! & (n+1)! & (n+2)!\\
(n+1)! & (n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)!
\end{pmatrix}$$
And $D=\det(A)$,
We have to prove that $D/(n!)^3 - 4$ is divisible by $n$.
I did it by simplifying the determinant using some row and column operatio... | $$
\begin{align}
D&=\det\begin{bmatrix}
n!&(n+1)!&(n+2)!\\
(n+1)!&(n+2)!&(n+3)!\\
(n+2)!&(n+3)!&(n+4)!
\end{bmatrix}\\
&=n!(n+1)!(n+2)!
\det\small\begin{bmatrix}
1&1&1\\
n+1&n+2&n+3\\
(n+1)(n+2)&(n+2)(n+3)&(n+3)(n+4)\\
\end{bmatrix}\tag{1}\\
&=n!(n+1)!(n+2)!
\det\begin{bmatrix}
1&1&1\\
0&1&2\\
0&2n+4&4n+10\\
\end{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Evaluate integral $\int_{0}^{2\pi}\frac{1}{1+\cos^2(\theta)}d\theta$ using contours I am trying to evaluate by using contour integration:
$$\int_{0}^{2\pi}\frac{1}{1+\cos^2(\theta)}d\theta$$
From "Fundamentals of Complex Analysis," I have the following two formulas:
$$d\theta = \frac{dz}{iz}, \cos(\theta)=\frac{1}{2}(... | The same approach but with only one residue to evaluate:
\begin{align*}
\int_{0}^{2\pi}\frac{d\theta}{1+\cos^2(\theta)}&=\int_{0}^{2\pi}\frac{2d\theta}{3+\cos(2\theta)}=\int_{0}^{4\pi}\frac{dt}{3+\cos(t)}=2\int_{0}^{2\pi}\frac{dt}{3+\cos(t)}\\&=\frac{4}{i}\int_{|z|=1}\frac{dz}{6z+z^2+1}
=8\pi\,\mbox{Res}\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.